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The space underlying the projective unitary group of a separable, infinite-dimensional Hilbert space has a number of topologies, so for the purposes of this question, pick you favourite and answer for that one.

I've read that $PU(H)$ is a Fréchet manifold, but that was without saying which topology. There are two ways I can see to think about this. First is that if we know what sort of manifold $U(H)$ is, then we know what $PU(H)$ is, as the former looks locally like a chart of $PU(H)$ times $U(1)$. So we can consider the closed (I think!) subspace $U(H) \subset End(H)$.

Alternately we can consider $PU(H)$ as sitting inside the Hilbert-Schmidt operators on $H$ as the projective unitaries act freely on the latter. (EDIT: this is not right, as pointed out by Andrew. I was thinking of the inclusion $PU(H) \hookrightarrow U(HS(H))$, which doesn't really tell us much in hindsight.)

I'm not familiar enough with the analysis to turn the above observations into results, and I may be interested in other topologies.

So the question is, is there the structure of a Banach or even Hilbert manifold on $PU(H)$?

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Check out math.ucr.edu/home/baez/calgary/BG.html beginning with note 3. –  Benjamin Dickman Nov 24 '12 at 10:16
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1 Answer

Edit: Theo's comments are spot-on - that'll teach me to try to post a quick answer without thinking too much

There are two sensible topologies on $U(H)$, each leads to a topology on $P U(H)$ by quotienting:

  1. The norm topology. In this topology, $U(H)$ is viewed as a sub-Lie group of the Banach Lie group $Gl(H)$. Here, $Gl(H)$ is open in $B(H)$ and the exponential map is a diffeomorphism onto a neighbourhood of the identity and this takes skew-Hermitian operators onto $U(H)$, thus defining charts. As the quotient $U(H) \to P U(H)$ admits slices, this makes $P U(H)$ a Banach manifold.

  2. The weak topology. In this topology, we start with $Gl(H)$ as a subspace of $B(H)$ where this has the weak topology in which $A_\lambda \to A$ if $A_\lambda v \to A v$ for each $v \in H$. However, this isn't good enough for $Gl(H)$ as taking the inverse isn't continuous. So we put on it the topology so that both the inclusion $Gl(H) \to B(H)$ and the inclusion after the inversion map $A \to A^{-1}$ are continuous. However, then $Gl(H)$ is not an open subspace of $B(H)$ so it isn't obviously a manifold. Indeed, I don't think that with this topology then it is a manifold at all (I don't think it is an ANR). As Theo says, the induced topology on $U(H)$ doesn't need the inversion fix, but even so it still is not a manifold.

The map to Hilbert-Schmidt operators is not useful here because the action is not free, it is only faithful. It acts by conjugation whereupon the stabiliser group of a particular element are all the unitary elements that commute with it and this can be quite large.

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I don't understand your answer: Ad 1) $U(H)$ is not open in $GL(H)$ since it doesn't contain any norm ball around the identity. How is $GL(H)$ a Banach space? It's a non-trivial open subset of $B(H)$. In order to show that $U(H)$ is a Lie group I'd try using the implicit function theorem. Ad 2) You describe the strong operator topology which happens to coincide with the weak operator topology on $U(H)$ (where this is Polish for separable $H$). Taking the adjoint is strongly cont. on normal operators, and multiplication is strongly cont. on norm-bounded sets, so this is a group topology on U(H) –  Theo Buehler Nov 25 '12 at 3:18
To be a bit more constructive: For 1) I'd expect that the exponential map should model $U(H)$ on the skew-hermitian operators in $B(H)$. For 2) I'd expect that for separable $H$ the Stone representation theorem for strongly continuous one-parameter unitary groups via unbounded self-adjoint operators should be of use although I do not see how this would give rise to a manifold structure. –  Theo Buehler Nov 25 '12 at 3:44
Theo, you're absolutely right. I was thinking "$Gl(H)$" and writing quickly. I've done a quick-fix. –  Andrew Stacey Nov 25 '12 at 10:01
The weak topology on $U(H)$ and $PU(H)$ has one very nice property: in that topology $U(H)$ and $PU(H)$ are Polish groups. As a consequence, many other topologies (strong, ultraweak, etc.) end up being equal to the weak topology. The reason is that there is a general theorem about Polish groups that says that a continuous bijective homomorphism between Polish groups is necessarily a homeomorphism. –  André Henriques Nov 25 '12 at 19:47
Thanks for clarifying and correcting, Andrew. I suspected something to that effect -- this happens ot all of us :) (I leave my comments up for now because your answer refers to them. Feel free to edit the references to my comments out of your answer, and I'd remove the comments). // The point of my mentioning "Polish" in my first comment was precisely what André said. My favorite reference for this circle of ideas is Kechris's Classical Descriptive Set Theory, GTM 156, the general theorem André mentions being (12.24) on page 81. –  Theo Buehler Nov 25 '12 at 20:35
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